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TS EAMCET 9 May 2024 Shift 1 Solved Paper

Section: Mathematics

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Question : 6 of 160

Marks: +1, -0

\det \begin{vmatrix} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{vmatrix} =

[TG EAPCET 9 May 2024 Shift 1]

$\& (a-b)(b-c)(c-a)$
$(a+b)(b+c)(c+a)$
$2 a b c$
$4 a b c$

Solution:

\text{ Let } \; \Delta = \begin{vmatrix} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{vmatrix}

On multiply

R_1, R_2

and

R_3

by

c, a

and

b

, respectively

\Rightarrow \Delta = \frac{1}{abc} \begin{vmatrix} a^2+b^2 & c^2 & c^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix}

Apply

R_1 \rightarrow R_1- (R_2+R_3)

\therefore \Delta = \frac{1}{abc} \begin{vmatrix} 0 & -2 b^2 & -2 a^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix}

\;=\; \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix}

Apply

R_2 \rightarrow R_2-R_1

and

R_3 \rightarrow R_3-R_1

, we get

\;=\; \frac{-2}{abc} \begin{vmatrix} 0 & b^2 & a^2 \\ a^2 & c^2 & 0 \\ b^2 & 0 & c^2 \end{vmatrix}

\;=\; \frac{-2}{abc} \left[ 0-b^2 (a^2 c^2-0) +a^2 (0-b^2 c^2) \right]

\;=\; \frac{-2}{abc} (-2 a^2 b^2 c^2) =4 a b c

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